Question 58593
The one formula that I'm going to use that you may not remember for factoroing the difference of cubes is {{{highlight(u^3-v^3=(u-v)(u^2+uv+v^2))}}}
{{{(x^3-1)/(x^2+1)}}} divided by {{{(9x^2+9x+9)/(x^2-x)}}}
First flip the second fraction and multiply:
{{{((x^3-1)/(x^2+1))*((x^2-x)/(9x^2+9x+9))}}}  Now factor everything:
{{{((x-1)(x^2+x+1)/(x^2+1))*(x(x-1)/(9(x^2+x+1)))}}}  Cancel the numerators and denominators that match:
{{{((x-1)*cross((x^2+x+1))/(x^2+1))*(x(x-1)/(9*cross((x^2+x+1))))}}}  Multiply whats left:
{{{((x-1)*x(x-1))/((x^2+1)*9)}}}
{{{x(x-1)^2/(9(x^2+1))}}}
Most books and teachers are happy with this, though you can multiply everything through.
Happy Calculating!!!